Can you solve this? The concentration of the #Zn^(2+)# ions in blood serum is 1.2 ppm. Express this amount in terms of millimole/L and milliequivalent/L.

1 Answer
Jul 10, 2015

Answer:

The concentrations are #1.9 * 10^(-2)"mM"# and #3.8 * 10^(-2)"meq/L"#.

Explanation:

So, you know that the ppm concentration of zinc cations, #Zn^(2+)#, in blood serum is equal to 1.2 ppm.

A 1 ppm concentration means that you have 1 part solute, in your case zinc cations, per 1 million parts of solvent.

To get the ppm concentration, simply multiply the ratio that exists between the mass of the solute and the mass of the solvent by 1 million, or #10^6#.

To make calculations easier, assume that you're dealing with a 1-L sample of blood serum. The density of blood serum is know to be equal to #"1.025 g/mL"#, which means that your liter of serum will have a mass of

#1cancel("L") * (1000cancel("mL"))/(1cancel("L")) * "1.025 g"/(1cancel("mL")) = "1.025 g"#

Use the definition of ppm concentration to determine what mass of zinc cations you have in solution

#"ppm" = m_"solute"/m_"solvent" * 10^6#

#m_"solute" = ("ppm" * m_"solvent")/10^6#

#m_"solute" = (1.2 * "1025 g")/10^6 = 1.23 * 10^(-3)"g"#

To express the concentration in milimoles per liter, you need to determine how many moles of #Zn^(2+)# you have in the sample. To do that, use its molar mass

#1.23 * 10^(-3)cancel("g") * "1 mole"/(65.39cancel("g")) = 1.9 * 10^(-5)"moles"# #Zn^(2+)#

Expressed in milimoles, this is equal to

#1.9 * 10^(-5)cancel("moles") * "1000 mmoles"/(1cancel("mole")) = 1.9 * 10^(-2)"mmoles"# #Zn^(2+)#

The concentration will thus be

#C = n/V = (1.9 * 10^(-2)"mmoles")/("1 L") = color(green)(1.9 * 10^(-2)"mM")#

To determine the concentration in miliequivalent per liter, you need to take into consideration the fact that normality only makes sense in the context of a chemical reaction.

An equivalent is simply a reactive unit - in your case, a reactive unit is a monovalent cation.

Since the zinc cation is actually a divalent cation, the number of equivalents will be equal to 2.

This means that you have

#N = "eq" * C#, where

#"eq"# - the number of equivalents;
#C# - the molarity of the solution.

Plug in your values to get

#N = 2 * 1.9 * 10^(-2)"mM" = color(green)(3.8 * 10^(-2)"meq/L")#

SIDE NOTE I think that you can actually use the ppm concentration per liter of solution, which would make the mass of the zinc cations equal to

#m_"solute" = (1.2 * 1000)/10^6 = 1.2 * 10^(-3)"g"#

As you can see, the values are not that different.